$ A = \left[\begin{array}{rr}-1 & 3 \\ 2 & 1\end{array}\right]$ $ C = \left[\begin{array}{rr}-2 & 4 \\ 1 & 4\end{array}\right]$ What is $ A C$ ?
Explanation: Because $ A$ has dimensions $(2\times2)$ and $ C$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ A C = \left[\begin{array}{rr}{-1} & {3} \\ {2} & {1}\end{array}\right] \left[\begin{array}{rr}{-2} & \color{#DF0030}{4} \\ {1} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ A$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ A$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ A$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{-2}+{3}\cdot{1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ A$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{-2}+{3}\cdot{1} & ? \\ {2}\cdot{-2}+{1}\cdot{1} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ A$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{-2}+{3}\cdot{1} & {-1}\cdot\color{#DF0030}{4}+{3}\cdot\color{#DF0030}{4} \\ {2}\cdot{-2}+{1}\cdot{1} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{-2}+{3}\cdot{1} & {-1}\cdot\color{#DF0030}{4}+{3}\cdot\color{#DF0030}{4} \\ {2}\cdot{-2}+{1}\cdot{1} & {2}\cdot\color{#DF0030}{4}+{1}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}5 & 8 \\ -3 & 12\end{array}\right] $